Consider a locally compact group $G$, considered as a measurable space with the completed Borelstructure wrt. the Haarmeasure. Consider a map $f:G \to G$, which is measurable and has an inverse, which is then also measurable. Is $f$ an homeomorphism?

What if $G$ is abelian? If not, what are necessary conditions on $G$, such that this is the case.

Surely you mean that $f$ is a group homomorphism? Could you also clarify that you really do mean "measurable" in this strong sense?
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Matthew DawsMay 6 '11 at 13:17

If you are interested in a fixed measure, then the completion of the $\sigma$ algebra wrt. to this measure means that you consider the $\sigma$ algebra generated by the measurable sets and the set of outer zero measure, or not? I thought that might simplify a lot, but I am not really experienced in measure theory.
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Marc PalmMay 6 '11 at 13:42

For example, as Robin Chapman says: mathoverflow.net/questions/19402/… the map $\mathbb R \rightarrow \mathbb R^2; x\mapsto (x,0)$ is continuous (and so Borel measurable), even a group homomorphism, but it is not measurable if both $\mathbb R$ and $\mathbb R^2$ are given their completed sigma-algebras for Lebesgue measure.
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Matthew DawsMay 6 '11 at 14:00

4 Answers
4

Here is a result by Adam Kleppner (MEASURABLE HOMOMORPHISMS OF LOCALLY COMPACT GROUPS, PROC AMER MATH SOC , vol. 106, no. 2, 1989, 391-395): any measurable homomorphism between locally compact groups is continuous. Actually what he really needs, for a homomorphism $\alpha:G\rightarrow H$, is that $\alpha^{-1}(U)$ is measurable in $G$ for every open subset $U\subset H$.

The basic fact about locally compact groups is that you can recover the topology from the underlying measure space:

This is because, for any measurable subset $X\subset G$ of positive measure, the set
$$
X^{-1}X:=\{x^{-1}y\\,|\\,x\in X, y\in Y\}
$$
is a neighborhood of the neutral element.
Letting $X$ vary along all measurable subset of positive measure you get a basis of neighborhoods of $e\in G$. By translating by group elements, you get a basis of neighborhoods of any element $g\in G$. And so you recover the topology on $G$.

Corollary:
Since the topology is entirely encoded in the measurable structure, an automorphism that respects the measurable structure, will also respect the topology, i.e., be continuous.

I'd like to note here that the statement that $X^{-1}X$ contains an open neighborhood of $e$ is not so hard to prove if one notes that $f(y)=\int \chi_{X}(x)\chi_{X}(xy^{-1}) dx$ is continuous (replacing $X$ by a set of finite measure it contains) and consider what happens when $y=0.$
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Benjamin HayesMay 7 '11 at 11:57

Some partial results: By Hewitt+Ross, Theorem 22.18 (http://www.ams.org/mathscinet-getitem?mr=551496) a Borel measurable homomorphism between two locally compact groups is continuous if the codomain is separable or $\sigma$-compact. (I think this result goes back to Banach).

I'm afraid that I don't know the limits of these sort of results (i.e. a counter-example in the non-separable case, say), or if being an automorphisms gives anything more.

Edit: As Julien points out in a comment, this paper of Neeb is a little suspect, so I withdraw my recommendation. André Henriques shows, in a short argument, that given a bijective group homomorphism which is measurable for the completed Haar measure, the homomorphism must be continuous.

I was a bit worried about the difference between "measurable" in the sense of "inverse image of open set is Borel or Haar measurable" and this stronger sense. But I think uniqueness of Haar measures rescues us. Indeed, if $\tau:G\rightarrow G$ is a continuous automorphism of $G$, then the map $A\mapsto |\tau(A)|$ will be a left invariant measure; as $\tau$ is a homeomorphism, this measure also assigns finite measure to compacts, and non-zero measure to open sets. Thus it will be proportional to the Haar measure. As $\tau$ preserves Borel sets, it follows that it will preserve all the Haar measurable sets, and so will be measurable in this strong sense. Note that the example of Robin Chapman shows that this isn't necessarily so for a merely injective, continuous homomorphism.

I should say that this refers to "Borel set" not "Completed Borel for Haar measure". The latter seems like a somewhat stronger condition, a priori.
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Matthew DawsMay 6 '11 at 12:43

4

The basic thing in the proof: if $A$ is a measurable set with postive measure, then $A A^{-1}$ contains a neighborhood of the identity.
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Gerald EdgarMay 6 '11 at 13:09

4

This paper by Neeb is actually not so nice - first, it seems to ignore all the literature on the subject (there are simpler proofs of that theorem of Banach, without the unnecessary assumptions on arcwise connectedness, and they were available before publication of his paper; actually, the result appears in textbooks, like Kechris' book); second, there is a major error in the part about representation theory (claiming that the operator norm topology and strong operator topologies have the same Borel sets, which is dead wrong). There seems to be an erratum of sorts for that paper (MR1747686 ).
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Julien MellerayMay 8 '11 at 15:14

@Julien: Thanks for the heads-up about the erratum! Yeah, I coming to the same conclusion (given the much better papers listed above).
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Matthew DawsMay 8 '11 at 19:06

2

Actually, there is a further major mistake: the reason for the existence of the paper seems to be a confusion two completely different meanings of "Baire sets" (the one Banach uses is the $\sigma$-algebra generated by the Borel sets and the meager sets) and the other one is the $\sigma$-algebra by the compact $G_{\delta}$'s. I give a reference to Banach's work in my answer here mathoverflow.net/questions/57616/… and François's answer to the same question points to Pettis' standard results Julien quotes.
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Theo BuehlerMay 8 '11 at 19:29